Olympic SV 1999

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6th INTERNATIONAL COMPETITION FOR UNIVERSITY
STUDENTS IN MATHEMATICS

Keszthely, 1999.

Problems and solutions on the first day

1. _{a) Show that for any} _m_∈N _{there exists a real} _m_×_m_matrix_A _{such that}_A3 ₌_A₊_I_{, where} _I _{is the}

m×midentity matrix. (6 points)

b) Show that detA >0 for every realm×mmatrix satisfyingA3=A+I. (14 points)
Solution. _{a) The diagonal matrix}

A=λI =




λ 0

. ..

0 λ




is a solution for equationA3₌_A₊_I _{if and only if}_λ3 ₌_λ_{+ 1, because}_A3₋_A₋_I _{= (}_λ3₋_λ₋₁₎_I_{. This}
equation, being cubic, has real solution.

b) It is easy to check that the polynomialp(x) =x3₋_x_{−1 has a positive real root}_λ

1(becausep(0)<0)
and two conjugated complex rootsλ2 andλ3 (one can check the discriminant of the polynomial, which is

−1
3

3
+ −1

2

2

= ₁₀₈23 >0, or the local minimum and maximum of the polynomial).

If a matrix A satisfies equation A3 ₌ _A₊_I_{, then its eigenvalues can be only} _λ_1, _λ

2 and λ3. The
multiplicity ofλ2andλ3must be the same, becauseAis a real matrix and its characteristic polynomial has
only real coefficients. Denoting the multiplicity ofλ1 byαand the common multiplicity ofλ2andλ3 byβ,

detA=λα
1λ

β
2λ

β

3 =λα1 ·(λ2λ3)β.

Becauseλ1 andλ2λ3=|λ2|2 are positive, the product on the right side has only positive factors.
2. _{Does there exist a bijective map}_π_:N_→N _{such that}

∞
X

n=1

π(n)

n2 <∞?
(20 points)

Solution 1. _{No. For, let}_π _{be a permutation of}N_{and let}_N _∈N_{. We shall argue that}
3N

X

n=N+1

π(n)

n2 >
1
9.

In fact, of the 2N numbersπ(N + 1), . . . , π(3N) only N can be ≤N so that at leastN of them are > N.

Hence

3N

X

n=N+1

π(n)

n2 ≥
1
(3N)2

3N

X

n=N+1

π(n)> 1

9N2 ·N·N =
1
9.

Solution 2. _Let_π _{be a permutation of}N_{. For any}_n_∈N_{, the numbers}_π₍₁₎_{, . . . , π}₍_n_{) are distinct positive}
integers, thusπ(1) +. . .+π(n)≥1 +. . .+n= n(n+1)₂ . By this inequality,

∞

X

n=1

π(n)

n2 =

∞
X

n=1

π(1) +. . .+π(n)

1

n2 −
1
(n+ 1)2

≥

≥

∞

X

n=1

n(n+ 1)

2 ·

2n+ 1

n2₍_n_{+ 1)}2 =

∞
X

n=1

2n+ 1
2n(n+ 1) ≥

∞
X

n=1
1

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6th INTERNATIONAL COMPETITION FOR UNIVERSITY
STUDENTS IN MATHEMATICS

Keszthely, 1999.

Problems and solutions on the second day

1. _{Suppose that in a not necessarily commutative ring} _R _{the square of any element is 0. Prove that}

abc+abc= 0 for any three elementsa, b, c. (20 points)

Solution. _{From 0 = (}_a₊_b₎2 ₌_a2₊_b2₊_ab₊_ba₌_ab₊_ba_{, we have}_ab₌₋₍_ba_{) for arbitrary} _{a, b}_{, which}
implies

abc=a(bc) =− (bc)a=− b(ca)= (ca)b=c(ab) =− (ab)c=−abc.

2. _{We throw a dice (which selects one of the numbers 1}_,₂_{, . . . ,}_{6 with equal probability)}_n_{times. What is}
the probability that the sum of the values is divisible by 5? (20 points)

Solution 1. _{For all nonnegative integers} _n _{and modulo 5 residue class} _r_{, denote by} _p(r)_n _{the probability}
that after n throwing the sum of values is congruent to r modulo n. It is obvious that p(0)0 = 1 and

p(1)₀ =p(2)₀ =p(3)₀ =p(4)₀ = 0.

Moreover, for anyn >0 we have

p(r)
n =

6

X

i=1

1
6p

(r−i)

n−1 . (1)

From this recursion we can compute the probabilities for small values of nand can conjecture that p(r)n =
1

5 +
4

5·6n if n ≡ r (mod )5 and p
(r)

n = 1₅ − ₅_·1₆n otherwise. From (1), this conjecture can be proved by
induction.

Solution 2. _Let_S _{be the set of all sequences consisting of digits 1}_{, . . . ,}_{6 of length}_n_{. We create collections}
of these sequences.

Let a collection contain sequences of the form
66. . .6

| {z }

k

XY1. . . Yn−k−1,

where X ∈ {1,2,3,4,5}and k and the digits Y1, . . . , Yn−k−1 are fixed. Then each collection consists of 5
sequences, and the sums of the digits of sequences give a whole residue system mod 5.

Except for the sequence 66. . .6, each sequence is the element of one collection. This means that the
number of the sequences, which have a sum of digits divisible by 5, is 1

5(6

n₋_{1) + 1 if} _n _{is divisible by 5,}
otherwise 1₅(6n₋_1).

Thus, the probability is 1₅+₅_·4₆n ifnis divisible by 5, otherwise it is
1
5−

1
5·6n.

Solution 3. _{For arbitrary positive integer}_k_{denote by}_p_k _{the probability that the sum of values is}_k_{. Define}
the generating function

f(x) =

∞
X

k=1

pkxk =

x+x2₊_x3₊_x4₊_x5₊_x6
6

n

.

(The last equality can be easily proved by induction.)
Our goal is to compute the sum

∞
P

k=1

p5k. Letε= cos2π₅ +isin2π₅ be the first 5th root of unity. Then

∞
X

k=1

p5k= f(1) +f(ε) +f(ε

2_{) +}_f₍_ε3_{) +}_f₍_ε4₎

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Obviously f(1) = 1, and f(εj_{) =} εjn

6n forj = 1,2,3,4. This implies thatf(ε) +f(ε

2_{) +}_f₍_ε3_{) +}_f₍_ε4₎
is 4

6n ifnis divisible by 5, otherwise it is

−1
6n. Thus,

∞
P

k=1

p5k is 1₅+₅_·4₆n ifnis divisible by 5, otherwise it is
1

5−
1
5·6n.

3. _{Assume that}_x₁_{, . . . , x}_n_{≥ −1 and}
n

P

i=1

x3

i = 0. Prove that
n

P

i=1

xi≤ n₃. (20 points)

Solution. _{The inequality}

0≤x3−3
4x+

1

4 = (x+ 1)

x−1

2

2

holds forx≥ −1.

Substituting x1, . . . , xn, we obtain

0≤
n

X

i=1

x3i −

3
4xi+

1
4

=
n

X

i=1

x3i −
3
4

n

X

i=1

xi+

n

4 = 0−
3
4

n

X

i=1

xi+

n

4,

so Pn
i=1

xi≤n₃.

Remark. _{Equailty holds only in the case when}_n_{= 9}_k_,_k_{of the}_x₁_{, ..., x}_n _are_{−1, and 8}_k _{of them are} 1

2.
4. _{Prove that there exists no function}_f _{: (0}_,_+∞)_→₍₀_,_{+∞) such that}_f2₍_x₎_≥_f₍_x₊_y₎ _f₍_x_{) +}_y_{for any}

x, y >0. (20 points)

Solution. _{Assume that such a function exists. The initial inequality can be written in the form} _f₍_x₎₋

f(x+y)≥f(x)−_f(x)+yf2(x) = _f(x)+yf(x)y .Obviously,f is a decreasing function. Fixx >0 and choosen∈N_such
thatnf(x+ 1)≥1.Fork= 0,1, . . . , n−1 we have

f

x+k

n

−f

x+k+ 1

n

≥ f x+

k
n

nf x+ k

n

+ 1 ≥
1
2n.

The additon of these inequalities givesf(x+ 1)≤f(x)−1₂. From this it follows thatf(x+ 2m)≤f(x)−m

for allm∈N_._Taking_m_≥_f₍_x_{), we get a contradiction with the conditon}_f₍_x₎_>₀_.

5. _Let_S_{be the set of all words consisting of the letters}_{x, y, z}_{, and consider an equivalence relation}_∼_on_S
satisfying the following conditions: for arbitrary wordsu, v, w∈S

(i)uu∼u;

(ii) ifv∼w, thenuv∼uwandvu∼wu.

Show that every word inS is equivalent to a word of length at most 8. (20 points)

Solution. _{First we prove the following lemma: If a word} _u_∈ _S _{contains at least one of each letter, and}

v∈S is an arbitrary word, then there exists a wordw∈S such thatuvw∼u.

If v contains a single letter, sayx, write uin the form u=u1xu2, and choosew =u2. Then uvw=
(u1xu2)xu2=u1((xu2)(xu2))∼u1(xu2) =u.

In the general case, let the letters ofv bea1, . . . , ak. Then one can choose some wordsw1, . . . , wk such
that (ua1)w1 ∼u, (ua1a2)w2 ∼ua1, . . ., (ua1. . . ak)wk ∼ua1. . . ak−1. Thenu∼ua1w1 ∼ua1a2w2w1 ∼

. . .∼ua1. . . akwk. . . w1=uv(wk. . . w1), sow=wk. . . w1 is a good choice.

Consider now an arbitrary worda, which contains more than 8 digits. We shall prove that there is a
shorter word which is equivalent to a. Ifa can be written in the formuvvw, its length can be reduced by

uvvw∼uvw. So we can assume thatadoes not have this form.

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It is easy to check thatb anddcontains all the three lettersx,y andz, otherwise their length could be
reduced. By the lemma there is a worde such that b(cd)e∼b, and there is a word f such that def ∼d.
Then we can write

a=bcd∼bc(def)∼bc(dedef) = (bcde)(def)∼bd.

Remark. _{Of course, it is enough to give for every word of length 9 an shortest shorter word. Assuming that}

the first letter isxand the second isy, it is easy (but a little long) to check that there are 18 words of length
9 which cannot be written in the formuvvw.

For five of these words there is a 2-step solution, for example

xyxzyzx zy∼xy xzyz xzyzy∼xyx zy zy∼xyxzy.

In the remaining 13 cases we need more steps. The general algorithm given by the Solution works

for these cases as well, but needs also very long words. For example, to reduce the length of the word

a=xyzyxzxyz, we have setb=xyzy,c=x,d=zxyz, e=xyxzxzyxyzy,f =zyxyxzyxzxzxzxyxyzxyz.
The longest word in the algorithm was

bcdedef=xyzyxzxyzxyxzxzyxyzyzxyzxyxzxzyxyzyzyxyxzyxzxzxzxyxyzxyz,

which is of length 46. This is not the shortest way: reducing the length of wordacan be done for example
by the following steps:

xyzyxzx yz ∼xyzyxz xyzy z∼xyzyxzxy zyx yzyz∼xyzyxz xyzyxz yx yz yz∼xy zyx zyx yz∼xyzyxyz.

(The last example is due to Nayden Kambouchev from Sofia University.)
6. _Let_A_{be a subset of}Z_n ₌Z_/nZ _{containing at most} 1

100lnnelements. Define therth Fourier coefficient
ofAforr∈Z_n _by

f(r) =X
s∈A

exp

2πi
n sr

.

Prove that there exists anr6= 0, such that f(r)≥ |A₂|. (20 points)
Solution. _Let_A₌_{_a₁_{, . . . , a}_{k}. Consider the}_k_-tuples

exp2πia1t

n , . . . ,exp

2πiakt

n

∈Ck_, _t_{= 0}_,₁_{, . . . , n}₋₁_.

Each component is in the unit circle|z|= 1. Split the circle into 6 equal arcs. This induces a decomposition
of the k-tuples into 6k _{classes. By the condition}_k_≤ 1

100lnnwe have n >6k, so there are two k-tuples in
the same class say fort1< t2. Set r=t2−t1. Then

Re exp2πiajr

n = cos

2πajt2

n −

2πajt1

n

≥cosπ
3 =

1
2
for allj, so

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3. _{Suppose that a function}_f _:R_→R_{satisfies the inequality}

n

X

k=1

3k f(x+ky)−f(x−ky)

≤1 (1)

for every positive integernand for allx, y∈R_{. Prove that}_f _{is a constant function. (20 points)}
Solution. _{Writing (1) with}_n₋_{1 instead of}_n_,

n−1

X

k=1

3k f(x+ky)−f(x−ky)

≤1. (2)

From the difference of (1) and (2),

3n f(x+ny)−f(x−ny)≤2;

which means

f(x+ny)−f(x−ny)≤ 2

3n. (3)

For arbitraryu, v ∈R_and_n_∈N_{one can choose}_x_and_y_{such that}_x₋_ny₌_u_and_x₊_ny₌_v_{, namely}

x=u+v

2 andy=
v−u

2n . Thus, (3) yields

f(u)−f(v)≤ 2
3n

for arbitrary positive integern. Because ₃2n can be arbitrary small, this impliesf(u) =f(v).
4. _{Find all strictly monotonic functions}_f _{: (0}_,_+∞)_→₍₀_,_{+∞) such that}_f x2

f(x)

≡x. (20 points)
Solution. _Let_g₍_x_{) =} f(x)

x .We haveg(
x

g(x)) =g(x). By induction it follows thatg(

x

gn₍_x₎) =g(x),i.e.

(1) f( x

gn₍_x₎) =

x

gn−1₍_x₎, n∈N.

On the other hand, let substitutexbyf(x) inf( x
2

f(x)) =x.¿From the injectivity off we get

f2₍_x₎

f(f(x))=

x,i.e. g(xg(x)) =g(x). Again by induction we deduce thatg(xgn₍_x_{)) =}_g₍_x_{) which can be written in the}
form

(2) f(xgn₍_x_{)) =}_xgn−1₍_x₎_{, n}_∈_N_.
Setf(m)₌_f_◦_f_◦_{. . .}_◦_f

| {z }

m times

.It follows from (1) and (2) that

(3) f(m)(xgn(x)) =xgn−m₍_x₎_{, m, n}_∈_N_.

Now, we shall prove that g is a constant. Assume g(x1) < g(x2). Then we may find n ∈ N _such
that x1gn(x1)≤x2gn(x2). On the other hand, ifmis even then f(m) is strictly increasing and from (3) it
follows thatxm

1gn

−m₍_x₁₎_≤_xm
2gn

−m₍_x₂₎_._{But when} _n_{is fixed the opposite inequality holds}_∀_m₁_. _This
contradiction shows thatg is a constant, i.e. f(x) =Cx, C >0.

Conversely, it is easy to check that the functions of this type verify the conditions of the problem.
5. _{Suppose that 2}_n_{points of an}_n_×_n_{grid are marked. Show that for some}_{k >}_{1 one can select 2}_k_distinct
marked points, saya1, . . . , a2k, such thata1anda2 are in the same row,a2 anda3are in the same column,

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Solution 1. _{We prove the more general statement that if at least}_n₊_k_{points are marked in an}_n_×_k_grid,
then the required sequence of marked points can be selected.

If a row or a column contains at most one marked point, delete it. This decreasesn+kby 1 and the
number of the marked points by at most 1, so the condition remains true. Repeat this step until each row
and column contains at least two marked points. Note that the condition implies that there are at least two
marked points, so the whole set of marked points cannot be deleted.

We define a sequenceb1, b2, . . .of marked points. Letb1be an arbitrary marked point. For any positive
integern, letb2n be an other marked point in the row ofb2n−1 andb2n+1 be an other marked point in the
column ofb2n.

Letmbe the first index for whichbmis the same as one of the earlier points, saybm=bl,l < m.
Ifm−l is even, the line segmentsblbl+1,bl+1bl+2, ...,bm−1bl=bm−1bmare alternating horizontal and
vertical. So one can choose 2k=m−l, and (a1, . . . , a2k) = (bl, . . . , bm−1) or (a1, . . . , a2k) = (bl+1, . . . , bm)
iflis odd or even, respectively.

Ifm−lis odd, then the pointsbl=bm,bl+1 andbm−1 are in the same row/column. In this case chose
2k =m−l−1. Again, the line segments bl+1bl+2, bl+2bl+3, ..., bm−1bl+1 are alternating horizontal and
vertical and one can choose (a1, . . . , a2k) = (bl+1, . . . , bm−1) or (a1, . . . , a2k) = (bl+2, . . . , bm−1, bl+1) if l is
even or odd, respectively.

Solution 2. _{Define the graph}_G_{in the following way: Let the vertices of}_G_{be the rows and the columns of}

the grid. Connect a rowrand a columnc with an edge if the intersection point ofrandc is marked.

The graphGhas 2nvertices and 2nedges. As is well known, if a graph ofN vertices contains no circle,
it can have at mostN−1 edges. ThusGdoes contain a circle. A circle is an alternating sequence of rows
and columns, and the intersection of each neighbouring row and column is a marked point. The required
sequence consists of these intersection points.

6. _{a) For each 1}_{< p <}_∞_{find a constant}_c_p_<_∞_{for which the following statement holds: If}_f _{: [−1}_,_1]_→R
is a continuously differentiable function satisfyingf(1)> f(−1) and|f0

(y)| ≤1 for ally∈[−1,1], then there
is anx∈[−1,1] such thatf0

(x)>0 and|f(y)−f(x)| ≤cp f0(x)

1/p

|y−x|for ally∈[−1,1]. (10 points)
b) Does such a constant also exist forp= 1? (10 points)

Solution. _{(a) Let} _g₍_x_{) = max(0}_{, f}0

(x)). Then 0 < R−11f

0

(x)dx = R−11g(x)dx+

R1

−1(f

0

(x)−g(x))dx, so
we get R−11|f

0

(x)|dx = R−11g(x)dx+

R1

−1(g(x)−f

0

(x))dx < 2R−11g(x)dx. Fix p and c (to be determined
at the end). Given any t > 0, choose for every x such that g(x) > t an interval Ix = [x, y] such that
|f(y)−f(x)|> cg(x)1/p_|_y₋_x_|_{> ct}1/p_|_I_x|_{and choose disjoint}_I

xi that cover at least one third of the measure
of the set {g > t}. For I = SiIi we thus have ct1/p|I| ≤ R_If0(x)dx ≤ R

1

−1|f

0

(x)|dx < 2R−11g(x)dx; so
|{g > t}| ≤3|I|<(6/c)t−1/pR1

−1g(x)dx. Integrating the inequality, we get

R1

−1g(x)dx =

R1

0 |{g > t}|dt <
(6/c)p/(p−1)R−11g(x)dx; this is a contradiction e.g. forcp= (6p)/(p−1).

(b) No. Given c > 1, denote α= 1/c and choose 0 < ε < 1 such that ((1 +ε)/(2ε))−α _< ₁_/_{4. Let}

g: [−1,1]→[−1,1] be continuous, even,g(x) =−1 for|x| ≤εand 0≤g(x)< α((|x|+ε)/(2ε))−α−1_for_{ε <}
|x| ≤1 is chosen such thatR_ε1g(t)dt >−ε/2+R_ε1α((|x|+ε)/(2ε))−α−1_dt₌₋_ε/₂₊₂_ε_(1−((1+_ε₎_/₍₂_ε₎₎−α₎_{> ε}_.
Letf =R g(t)dt. Thenf(1)−f(−1)≥ −2ε+ 2R_ε1g(t)dt >0. Ifε < x <1 andy=−ε, then|f(x)−f(y)| ≥
2ε−R_εxg(t)dt ≥ 2ε−R_εxα((t+ε)/(2ε))−α−1 _{= 2}_ε₍₍_x₊_ε₎_/₍₂_ε₎₎−α _{> g}₍_x_)|_x₋_y_|_/α ₌ _f0

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